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Chapter 8: Problem 76

At the local playground a child on a swing has a speed of \(2.02 \mathrm{m} / \mathrm{s}\) when the swing is at its lowest point. (a) To what maximum vertical height does the child rise, assuming he sits still and "coasts"? Ignore air resistance. (b) How do your results change if the initial speed of the child is halved?

Short Answer

Expert verified
The child rises to 0.208 m with a speed of 2.02 m/s; if halved, the height is 0.052 m.

Step by step solution

01

Understand Energy Conservation

At its lowest point, the swing's kinetic energy is maximum and its potential energy is zero. At the highest point, the kinetic energy is zero and the potential energy is maximum. Therefore, using the conservation of energy principle, we equate the kinetic energy at the lowest point to the potential energy at the highest point.
02

Write the Energy Conservation Equation

The kinetic energy at the lowest point is given by \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the child and \( v \) is the speed (2.02 m/s). The potential energy at the highest point is \( PE = mgh \), where \( h \) is the height and \( g \) is the gravitational acceleration (9.81 m/s虏). Setting \( KE = PE \), we have: \[ \frac{1}{2} mv^2 = mgh \].
03

Simplify and Solve for Height

Cancel out the mass \( m \), and solve for \( h \) to find the maximum height: \[ h = \frac{v^2}{2g} \]. Substitute the given initial speed: \( v = 2.02 \) m/s, so: \[ h = \frac{(2.02)^2}{2 \times 9.81} \approx 0.208 \text{ m} \].
04

Analyze the Scenario with Halved Speed

If the speed is halved, the new speed is \( v = 1.01 \) m/s. Recompute \( h \) using the formula \( h = \frac{v^2}{2g} \): \[ h = \frac{(1.01)^2}{2 \times 9.81} \approx 0.052 \text{ m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy of motion. It plays a crucial role in understanding how objects like a swinging child gain and lose energy. When the child reaches the lowest point in the swing, this is where kinetic energy is at its maximum. Kinetic energy depends on two key factors: mass and velocity. The formula for kinetic energy is given by
  • \( KE = \frac{1}{2} mv^2 \)
This means kinetic energy increases with the square of the speed.
If the child's speed is halved, the kinetic energy doesn't simply halve鈥攊t's reduced to one-fourth. Why? Because of the squared speed component in the formula.
Thus, changes in speed have a significant impact on kinetic energy and are crucial for predicting how high the child will swing.
Potential Energy
Potential energy is the energy stored due to an object's position. For the child on the swing, potential energy is highest at the maximum height they reach during the swing. Here, movement stops briefly, indicating zero kinetic energy and maximum potential energy.The formula for potential energy is:
  • \( PE = mgh \)
  • \( m \) is the mass,
  • \( g \) is gravitational acceleration (9.81 m/s虏), and
  • \( h \) is the height.
The energy conversion from kinetic to potential energy illustrates the principle of conservation of energy. At the lowest point, kinetic converts into potential as the swing rises. Understanding this equation allows us to relate the child's initial speed to the highest point they reach.
This conservation highlights why the potential energy becomes maximum when kinetic energy is zero.
Gravitational Acceleration
Gravitational acceleration, denoted by \( g \), is a constant that reflects the acceleration due to Earth's gravity. On Earth's surface, this value is approximately \( 9.81 \text{ m/s}^2 \).
Gravitational acceleration is a key component when calculating potential energy. It ensures that regardless of mass, the acceleration due to gravity remains constant. In our swing scenario, when switching energy from kinetic to potential, gravitational acceleration mediates how height is determined.
  • In calculating maximum height, \( g \) influences how high the child can rise based on their speed.
Remember, the impact of \( g \) doesn't change with velocity changes. However, the speed, both initially and while coasting, directly affects the energy transformation, demonstrating the limitation gravity puts on motion.
Understanding \( g \) not only simplifies working on physics problems but also enriches comprehension of real-world physics phenomena.

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Most popular questions from this chapter

Predict/Explain When a ball of mass \(m\) is dropped from rest from a height \(h,\) its speed just before landing is \(v .\) Now, suppose a second ball of mass \(4 m\) is dropped from rest from a height \(h / 4 .\) (a) Just before ball 2 lands, is its speed \(4 v, 2 v, v, v / 2\) or \(v / 4\) ? (b) Choose the best explanation from among the following: I. The factors of 4 cancel; therefore, the landing speed is the same. II. The two balls land with the same kinetic energy; therefore, the ball of mass \(4 m\) has the speed \(v / 2\) III. Reducing the height by a factor of 4 reduces the speed by a factor of 4.

Taking a leap of faith, a bungee jumper steps off a platform and falls until the cord brings her to rest. Suppose you analyze this system by choosing \(y=0\) at the platform level, and your friend chooses \(y=0\) at ground level. (a) Is the jumper's initial potential energy in your calculation greater than, less than, or equal to the same quantity in your friend's calculation? Explain. (b) Is the change in the jumper's potential energy in your calculation greater than, less than, or equal to the same quantity in your friend's calculation? Explain.

\(A\) 15,800-kg truck is moving at \(12.0 \mathrm{m} / \mathrm{s}\) when it starts down a \(6.00^{\circ}\) incline in the Canadian Rockies. At the start of the descent the driver notices that the altitude is \(1630 \mathrm{m}\). When she reaches an altitude of \(1440 \mathrm{m},\) her speed is \(29.0 \mathrm{m} / \mathrm{s}\). Find the change in (a) the gravitational potential energy of the system and (b) the truck's kinetic energy. (c) Is the total mechanical energy of the system conserved? Explain.

You coast up a hill on your bicycle with decreasing speed. Your friend pedals up the hill with constant speed. (a) Ignoring friction, does the mechanical energy of the you-bike-Earth system increase, decrease, or stay the same? Explain. (b) Does the mechanical energy of the friend-bike-Earth system increase, decrease, or stay the same? Explain.

A rock is thrown vertically upward from the top of a cliff that is \(32 \mathrm{m}\) high. When it hits the ground at the base of the cliff, the rock has a speed of \(29 \mathrm{m} / \mathrm{s}\). Assuming that air resistance can be ignored, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.
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